If three real normals can be drawn from a poi | vedclass

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(B) Let the point on the $x$-axis be $(h, 0)$.The equation of a normal to the parabola $y^2 = 4ax$ with slope $m$ is $y = mx - 2am - am^3$.Since the n

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$x > 2a$

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(B) Let the point on the $x$-axis be $(h, 0)$.The equation of a normal to the parabola $y^2 = 4ax$ with slope $m$ is $y = mx - 2am - am^3$.Since the normal passes through $(h, 0)$,we have $0 = mh - 2am - am^3$.Assuming $m 
eq 0$,we get $h = 2a + am^2$,or $am^2 = h - 2a$.For three real normals,the cubic equation in $m$ must have three real roots.This occurs when the point $(h, 0)$ lies inside the evolute of the parabola,which implies $h > 2a$.Thus,the $x$-coordinate must be greater than $2a$.

If three real normals can be drawn from a point on the $x$-axis to the parabola $y^2 = 4ax$ $(a > 0)$,what is the range of the $x$-coordinate of the point?

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